Introduction to Proofs

Transcription

1 Real Analysis Preview May 2014

2 Properties of R n Recall Oftentimes in multivariable calculus, we looked at properties of vectors in R n. If we were given vectors x =< x 1, x 2,, x n > and y =< y1, y 2,, y n >, then the dot product of x and y, denoted by x y, is given by x n y = x i y i Norms and Distance The norm of a vector x =< x 1, x 2,, x n >, denoted by x, is given by x = x x = n. We can find the distance between two points P = (p 1, p 2,, p n ) and Q = (q 1, q 2,, q n ), n denoted by P Q, is given by P Q = (p i q i ) 2 x 2 i

3 Theorems involving R n Cauchy-Schwarz Inequality Let x and y be vectors in R n. Then x y x y. The Triangle Inequality Let x and y be vectors in R n. Then x + y x + y.

4 Cauchy-Schwarz Inequality Proof. Notice for every real number t, we have (x i t + y i ) 2 0. Therefore n (x i t + y i ) 2 0 for every real number t. We can rewrite the second inequality in the form At 2 + Bt + C 0 where A = n xi 2, B = n x i y i, C = n y 2 i. If A > 0, then we let t = B A, which is a real number. By plugging in for all values of t, our inequality becomes B 2 AC 0, which is the desired inequality. The other case when A = 0 means B = 0. This leads to equality in the aforementioned inequality.

5 Triangle Inequality Proof. x + y = n (x i + y i ) 2 = n (x 2 i + 2x i y i + y 2 i ) = x 2 +2 x y + y 2 x 2 +2 x y + y 2 = ( x + y ) 2 where the second to last inequality follows from the Cauchy Schwarz Inequality

6 Applications of the Cauchy Schwarz Inequality Question Given that a, b, c, d, and e are real numbers such that a + b + c + d + e = 8 and a 2 + b 2 + c 2 + d 2 + e 2 = 16 determine the maximum value of e.

7 Applications of the Cauchy Schwarz Inequality Question Given that a, b, c, d, and e are real numbers such that a + b + c + d + e = 8 and a 2 + b 2 + c 2 + d 2 + e 2 = 16 determine the maximum value of e. Answer We can put the aforementioned equations into the form 8 e = a + b + c + d and 16 e 2 = a 2 + b 2 + c 2 + d 2. Using the Cauchy Schwarz Inequality we find a + b + c + d a 2 + b 2 + c 2 + d 2. By substituting we find (8 e) 2 4(16 e 2 ), which yields 0 e This occurs when a = b = c = d = 6 5.

8 Open n-ball Definition Let a be a given point in R n and let r be a given positive integer. The set of all points x R n such that x a < r is called an open n-ball of radius r and center a. We often denote this set by B(a; r). What will this look like in R, R 2, and R 3?

9 Open sets in R n Definition Let S be a subset of R n and assume a S. Then a is called an interior point of S if there is an open n-ball with center at a, all of whose points belong to S. Definition A set S in R n is called open if all its points are interior points. Example Give an example of open sets in R, R 2, and R 3. (For R 2 consider the set (a, b) (c, d)). Give an example of sets that are not open in R, R 2, and R 3.

10 Indexed Collection of Sets Definition Let {A α } α I be the collection of all the sets A α where α I. The union of the sets in {A α } α I is given by A α = {x x A α for some α I }. α I Definition Let {A α } α I be the collection of all the sets A α where α I. The union of the sets in {A α } α I is given by A α = {x x A α for every α I }. α I

11 Examples of Indexed Collection of Sets Example Consider the set A n = {x C x n 1 = 0} for n N. Describe the set n N A n. Let a R 3. Describe B(a; r) and B(a; r). r R + r R + Calculate ([ 1 n, 5 4 ] ) [ 2n, 10 n ]. n n N

12 Questions on Open Sets Unions of Open Sets Is the union of an indexed collection of open sets an open set? Intersections of Open Sets Is the intersection of an indexed collection of open sets an open set? Hint: Consider the intersection of the collection of sets A n = ( 1 n, 1 n ) for n N.

13 Theorem 1 Theorem The union of any collection of open sets is open.

14 Theorem 1 Theorem The union of any collection of open sets is open. Proof. Assume {A α } α I is a collection of open sets and let S = α I A α. Now select an arbitrary point x S. We must show that x is an interior point of S. Since x S, x must belong to at least one of the sets in {A α } α I. Without loss of generality, we can assume x A j where j I. Recall that A j is an open set, which implies there exists an open n-ball B(x; r) such that B(x; r) A j. Since A j S, we find B(x; r) S and hence x is an interior point of S. Therefore S is an open set.

15 Theorem 2 Theorem The intersection of a finite collection of open sets is open.

16 Theorem 2 Theorem The intersection of a finite collection of open sets is open. Proof. Let S = n A i where each A i is open. If S is empty, then S is open and we are finished. So we shall assume S is nonempty, and let x S. Then x A i for every i = 1, 2,, n and hence there is an open n-ball B(x; r i ) A i for every i = 1, 2,, n. Let r = min{r 1, r 2,, r n }. Then B(x; r) S implying x is an interior point. As a result, S is open.

17 Examples of Differences and Complements Example Let U = {a, b, c, d, e, f, g, h, i}, A = {a, d, e, f, h}, and B = {b, d, h, i}. Determine A B, B A, A, and B. If we are considering R as a subset of the complex numbers, how could we describe R. For all sets A and B does A B = A B.

18 Closed Sets Definition A set S in R n is called closed if its complement R n S is open. Example Give three examples of closed sets in R, R 2, and R 3. Is it possible for a set to be both closed and open? What can we say about the union and intersection of closed sets?

19 Theorem 3 Theorem The union of a finite collection of closed sets is closed.

20 Theorem 3 Theorem The union of a finite collection of closed sets is closed. Proof. Let S = n A i where each A i is closed. Consider R n S. Since R n S = R n ( S, we can apply De Morgan s Law to find n ) n n R n S = R n A i = (R n A i ) = (R n A i ). Since A i is closed, we find R n A i is open for i = 1, 2,, n. We know from Theorem 2 that a finite intersection of open sets is open. As a result, R n S is open and S is closed.

21 Theorem 4 Theorem The intersection of an arbitrary collection of closed sets is closed.

22 Theorem 4 Theorem The intersection of an arbitrary collection of closed sets is closed. Proof. Let S = i I A i where each A i is closed. Consider R n S. Since R n S = R n S, we can apply De Morgan s Law and distributive laws to find ( ) R n S = R n A i = (R n A i ) = i I i I i I(R n A i ). Since A i is closed, we find R n A i is open for all i I. We know from Theorem 1 that an arbitrary union of open sets is open. As a result, R n S is open and S is closed.

23 The Cantor Set Example We wish to construct a closed set using Theorem 4. We begin by deleting the open interval ( 1 3, 2 3) from the closed interval F 0 = [0, 1], and let F 1 denote the remaining closed set consisting of two closed intervals. We form the closed set F 2 consisting of four intervals by deleting the open intervals ( 1 9, 2 ( 9) and 7 9, 8 9) from F 1. We construct F 3 by deleting the middle third from each of the four closed intervals in F 2. Continue this process indefinitely to get a sequence of closed sets F n such that F 0 F 1 F 2 F n with the Cantor set being formed by the set given by F = We find F is closed by Theorem 4. F n. n=0

24 Properties of the Cantor Set What is left in the Cantor Set? Consider the numbers 0, 1, 1 3, 2 3, 1 9, 2 9, 7 9, 8 9,. Notice that all of the aforementioned numbers are in the Cantor set. Actually even more numbers exist in the Cantor set. Any number can be written in ternary form as a a a where a i = 0 or 2 for every i N. This leads us to an infinite number of elements in the Cantor set. In fact, an uncountable number of elements in the Cantor set.

25 More Properties of the Cantor Set Geometric Series Recall that a geometric series is of the form where a, r R with r < 1. ar k 1 = k=1 a 1 r

26 More Properties of the Cantor Set Geometric Series Recall that a geometric series is of the form where a, r R with r < 1. What is left in the Cantor Set? k=1 ar k 1 = k=1 a 1 r Notice that at the k th iteration of forming the Cantor set we have removed a total length of 2k 1 from the set [0, 1]. Thus the total 3 k length removed from [0, 1] is given by 2 k 1 3 k = 1 ( ) 2 k 1 = 1. This is somewhat astonishing 3 3 k=1 since the length of the intervals removed from [0, 1] equals the length of the total interval, yet an uncountable set remains in the Cantor set.